Question: Let $g(x)=5x^2+4\sin(x)$. $g'(x)=$
Solution: The expression for $g(x)$ includes $\sin(x)$. Remember that the derivative of $\sin(x)$ is $\cos(x)$. Put another way, $\dfrac{d}{dx}[\sin(x)]=\cos(x)$. $\begin{aligned} g'(x)&=\dfrac{d}{dx}[5x^2+4\sin(x)] \\\\ &=5\dfrac{d}{dx}(x^2)+4\dfrac{d}{dx}[\sin(x)] \\\\ &=5\cdot2x+4\cdot\cos(x) \\\\ &=10x+4\cos(x) \end{aligned}$ In conclusion, $g'(x)=10x+4\cos(x)$